I don't know how to draw in computers so I cannot show the sketch. But I can describe it.

f(x)=0,for x=0

That means the graph passes point (0,0).

f'(x)=0 for x=-2

That means the graph is horizontal at x = -2.

f"(x)=0 for x=1

That means the concavity of the graph changes at x = 1. There is an inflection point at x=1. The graph there needs not be horizontal, though.

f'(x)>0 for x<-2

The slope of the graph is positive to the left of x = -2.

So if at x = -2, the graph is horizontal, then there is maximum point at x = -2.

f'(x)<0 for -2<x<2

From x = -2 up to x = 2, the slope is negative.

f'(x)=0 for x>2

The graph is horizontal to the right of x = 2. Horizontal up to before infinity.

f"(x)<0 for x<1

The concavity of the graph is downwards to the left of x = 1.

That means from negative infinity up to x=1, the graph is like a vertical parabola that opens downward, whose vertex is at x = -2.

f" (x)>0 for 1<x<2

The concavity of the graph is upwards between x=1 and x=2. Yet, the slope there is still negative.

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Okay, if you are still with me, the graph is like a weak or tipsy number two.

From after negative infinity up to x=1 it looks like a parabola with its highest point at x = -2.

It passes through the origin (0,0), so the highest point of the "parabola" is above the x-axis.

Then at x=1 it changes concavity--it faces upward, but it is still going down.

Then at x=2 it becomes horizontal until it reaches very close to positive infinity.